証明|a|=|b|=|c|

lumea98

\[若a，b，c為整數，且\frac{a}{b}+\frac{b}{c}+\frac{c}{a}為整數\]
\[\frac{b}{a}+\frac{c}{b}+\frac{a}{c}亦為整數\]
\[証明|a|=|b|=|c|\]

求救~3Q

rrrrumia

$
abc|{a^2}c + a{b^2} + b{c^2},{b^2}c + a{c^2} + {a^2}b \\
a = \prod\limits_i {p_i^{{a_i}}} ,b = \prod\limits_i {p_i^{{b_i}}} ,c = \prod\limits_i {p_i^{{c_i}}}  \\
\prod\limits_i {p_i^{{a_i} + {b_i} + {c_i}}} |\prod\limits_i {p_i^{2{a_i} + {c_i}}}  + \prod\limits_i {p_i^{{a_i} + 2{b_i}}}  + \prod\limits_i {p_i^{{b_i} + 2{c_i}}} ,\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  + \prod\limits_i {p_i^{2{a_i} + {b_i}}}  + \prod\limits_i {p_i^{2{b_i} + {c_i}}}  \\
\forall k,p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {c_i}}}  + \prod\limits_i {p_i^{{a_i} + 2{b_i}}}  + \prod\limits_i {p_i^{{b_i} + 2{c_i}}} ,\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  + \prod\limits_i {p_i^{2{a_i} + {b_i}}}  + \prod\limits_i {p_i^{2{b_i} + {c_i}}}  \\
{a_k} \le {b_k} \le {c_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  + \prod\limits_i {p_i^{2{a_i} + {b_i}}}  + \prod\limits_i {p_i^{2{b_i} + {c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {b_i}}}  \Rightarrow {c_k} \le {a_k} \\
{a_k} \le {c_k} \le {b_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {c_i}}}  + \prod\limits_i {p_i^{{a_i} + 2{b_i}}}  + \prod\limits_i {p_i^{{b_i} + 2{c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {c_i}}}  \Rightarrow {b_k} \le {a_k} \\
{b_k} \le {a_k} \le {c_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {c_i}}}  + \prod\limits_i {p_i^{{a_i} + 2{b_i}}}  + \prod\limits_i {p_i^{{b_i} + 2{c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{a_i} + 2{b_i}}}  \Rightarrow {c_k} \le {b_k} \\
{b_k} \le {c_k} \le {a_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  + \prod\limits_i {p_i^{2{a_i} + {b_i}}}  + \prod\limits_i {p_i^{2{b_i} + {c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{b_i} + {c_i}}}  \Rightarrow {a_k} \le {b_k} \\
{c_k} \le {a_k} \le {b_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  + \prod\limits_i {p_i^{2{a_i} + {b_i}}}  + \prod\limits_i {p_i^{2{b_i} + {c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{a_i} + 2{c_i}}}  \Rightarrow {b_k} \le {c_k} \\
{c_k} \le {b_k} \le {a_k} \\
p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{2{a_i} + {c_i}}}  + \prod\limits_i {p_i^{{a_i} + 2{b_i}}}  + \prod\limits_i {p_i^{{b_i} + 2{c_i}}}  \Rightarrow p_k^{{a_k} + {b_k} + {c_k}}|\prod\limits_i {p_i^{{b_i} + 2{c_i}}}  \Rightarrow {a_k} \le {c_k} \\
$

isee

$
abc|{a^2}c + a{b^2} + b{c^2},{b^2}c + a{c^2} + {a^2}b \\
a = \prod\limits_i {p_i^{{a_i}}} ,b =  ...
rrrrumia 发表于 2016-10-21 14:20

mark 貌似搞定了，只是这符号不直接，看不太明表达，，，